The Carnot Cycle - Video Tutorials & Practice Problems

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1

concept

The Carnot Cycle and Maximum Theoretical Efficiency

Video duration:

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Hey guys. So in previous videos, we've talked about the second law of thermodynamics. And there was a statement that said that no heat engine could ever have an efficiency of 100%. But some problems are going to ask you to calculate something called the maximum theoretical efficiency between reservoirs. And to answer that, we're gonna look at something called the carnot cycle or the carnot engine. So this is something your professors and textbooks may make a big deal out of. What I'm going to show you this video is that it's really just a special type of cycle. And we're gonna see some equations that are really similar to ones that we've seen before. So let's check this out here. So basically the carnot cycle, some guy carnot back in the 1800s, he sat down, played around with a bunch of these diagrams and figured out that you could design an ideal reversible cycle and if you do this, it has the maximum possible efficiency. So basically what he was like, what he said was, well, if you can't get 100% how high can you actually go now, before I give you the equation here, I want to talk about this word reversible for just a second because you might see it. Get thrown around a lot and really anything, an engine or cycle is reversible basically. If it happens infinitely slowly, really, really, really slow and there's also no friction or anything like that. That takes away energy from the system. That's what reversible means. So let's take a look at this Carnot cycle here, it really just has four steps to ice a thermal and to idiomatic. So, the first one from A to B. Is an ice a thermal expansion. So, you're gonna ride this ice a therm down like this. And basically this is the step here where it's absorbing heat from the hot reservoir. So here is the step where it's taking in heat from the hot reservoir. Then what happens is that there's an idiomatic expansion. So, remember, that's gonna be steeper than an ice A therm. And it's gonna look like this from B to C. And what happens in this idiomatic is that there is no heat transfer, Right? That's the definition then, basically from the next, for the next two steps, it just gets reversed. So, here, now, there is an ice a thermal compression from C to D. It goes the opposite way like this. And in this step, what happens is that heat is flowing out to the cold reservoir. And then what happens is from D to A. There is an idiomatic compression, and then the cycle repeats itself. So, to a dramatic to ice, a thermal right? That's really what the carnot cycle is. And then the work that is done is just gonna be the area that's enclosed inside of this diagram. You don't really need to know that. But that's just the work. All right. So, what you might need to know though, is where the heat gets transferred in this cycle here. Now we just said that the steps two and four are idiomatic and what that means is that there's no heat transfer. So what that means is that it only gets transferred during steps one and three, the two ice of thermal ones. This is where you absorb heat and then release it. Now your textbooks are gonna have some pretty lengthy derivations, they're gonna do some equations with Mohler specific heats and acrobatics and all this and that. And basically what you're gonna see here is that the maximum possible efficiency is one minus TC over th notice how it looks very similar to the equation over here that we've been working with so far, it's just instead of queues were using teas, but that's basically what the maximum possible efficiency is. It depends only on the temperatures of the reservoirs themselves, that's what he found out. So, the the the the other equation that might need to know here is that the heat released and absorbed basically the ratio of Q. C. Over Q. H. Is equal to the absolute value of TC over th so, notice how all these are gonna be positive numbers when you do the absolute values. But that's really all that you need to know. You just need to know the maximum possible efficiency equation and then this one over here, that's really all you need, let's go ahead and take a look at our example. So, we have a carnot engine that is operating between 520 and 300. So we have T. H. Here is 520 T. C. Here is 300. So what happens is the engine is going to take in some amount of heat from the hot reservoir that's going to be Q. H. So this is gonna be 6.45 killer jewels. And then what happens is we want to calculate in part a what is the maximum theoretical efficiency? So what happens is E. Car? No this is gonna be the maximum theoretical efficiency. If this is a reversible engine is just gonna be one minus TC over T. H. So this is gonna be one minus and now we just use the heat or the temperatures of the hot and cold reservoirs. This is gonna be 300 divided by 5 20 And you're gonna get a theoretical maximum of 42%. So what this means here is that even if you could design the perfect engine that ran in this cycle here, the maximum efficiency that you could ever hope to even get out of this would be 42%. And that's the answer to part a. So part B. Now asks how much waste heat does the engine expel each cycle. We know Q. H. We don't know what work is but now we want to calculate what is Q. C. How do we calculate that? Well remember that? We have one equation. now that relates Qc with all of the other variables. So basically what happens is if we want Qc um we're gonna use Qc over Q. H. Absolute value is going to equal TC over th right the ratio of these heats is just for the ratio of the hot and cold reservoirs. So all you do here is Q. C. Is just gonna be equal to Q. H. Times TC over th we're just gonna use all positive numbers here. So there's gonna be 6.45 then this is gonna be um 300 divided by 5 20. When you work this out, what you're gonna get is 3.72 killer jewels. So this is going to be 3.72 killer jewels and that's the answer. Alright, so finally now let's take a look at part. See here, how much mechanical work does the engine produce? So we really where we want to calculate is this w here we have a couple of different ways to calculate this. Remember that W for the engine is always just uh the difference in Q. H minus Q. C. So we can actually just subtract these two right here. You can also just get it from the efficiency equation, there's a bunch of different ways but this is probably gonna be the most straightforward. So w is just gonna be um the heat that gets absorbed 6.45. Uh This is gonna be killer jewels minus 3.72 killer jewels, and then your answer is going to be 2.73 killer jewels. Alright, so basically, if you had a carnot cycle, this is as much work as you could possibly hope to extract per cycle. So that's it for this one. Guys, let me know if you have any questions.

2

Problem

Problem

A theoretical heat engine in space could operate between the Sun's 5500°C surface and the –270.3°C temperature of intergalactic space. What would be its maximum theoretical efficiency?

A

99.98%

B

95.1%

C

99.95%

3

Problem

Problem

A Carnot engine with an efficiency of 70% is cooled by water at 10°C. What temperature must the hot reservoir be maintained at?

A

33.3 K

B

404.3 K

C

943.3 K

D

14.3 K

4

example

Lifting Mass with a Heat Engine

Video duration:

6m

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Hey guys, So, I've got a great example here for you. We're gonna work this one out together. So we have this engine that's operating and it's extracting some heats and all this stuff, but we're asked to find out how many cycles it's gonna take to lift a 10 kg mass. So I want to draw out what's going on here. So imagine that I have some kind of a box that's on the ground like this, It's 10 kg and we want to lift it up by, you know, five m. So this height here is five. Well, we remember back from energy conservation, that if you want to lift a mass by some heights, you have to do some work. So, for example, the work that's required to lift this mass is equal to MG. H. And so therefore it's just 10 times 9.8 Times the five m. So in other words, it takes 490 jewels In order to lift this mass by five m. So, what does this have to do with the heat engine? What's going on here is I have a carnot engine that's operating between these temperatures and these heats, and that's the thing that's actually supplying the energy to lift this mass. So, I'm gonna draw out my sort of heat diagram or my, my energy flow diagram. So I've got my hot reservoir that's connected to my engine and this is connected to my cold reservoir, It's T. C. So what's happening here is this engine? This is my car, no engine here is actually the thing that is supplying the work in order to lift this mass. So, you can imagine that this carnot engine, like, you know, it's just like a box with a pulley or something, maybe it's like a little pulley and basically it's sort of cranking this this box To lift this mass by five m. So that's basically what's going on here. The thing that's supplying the work here is my car, no engine. And so remember, in a carnot cycle or in any heat engine cycle, you have a heat that flows in, you have some work that's done by the engine and then you have some heat that flows out to the hot the cold reservoir. So what happens here is we have to figure out how many cycles it takes for this engine. This work that's produced by the engine to lift this mass here. So, basically, what we're trying to find is the number of cycles times the work that's done in each cycle is going to equal the work that I need to lift this mass. So, basically, so this is going to be number of cycles, like this number of cycles times the work that I do per cycle Is going to equal the work that is required to lift this mass here. And ultimately, what I'm trying to find is what is in what is the number of cycles here. We know how much energy and how much work is required to lift, which is just the 490 that we calculated over here. So really all we have to do is figure out well how much work is produced by the engine each cycle. So basically that's really what we're trying to figure out in this part of the problem. So how do we calculate the work that's done by the engine? Well now we're just going to stick to our heat engine equations. We have a couple of them to deal with work and heats. So let's just try to use our sort of basic work equation. So W equals Q. H minus Q. C. So we do we have the Qh well and this probably we're told one of the heats these are the two temperatures 1 82 and zero. But we're told that it extracts 25 jewels of energy from the hot reservoir. So that's my cue. H this is gonna be 25 while I'm at it I'm just gonna fill in the rest of the variables here, my th this hot reservoir here is 1 82. So if you add to 73 to it, it's just going to be 455 Kelvin. We do the same thing, this is just gonna be T. C. Equals zero. So that's just gonna be 2 73 kelvin. Alright, so we don't have where we do have with the Q. H. Is unfortunately we don't have what Q. C. Is, what is the heat that gets discarded out to the cold reservoir. And we can't find it using this equation because we don't know W or Q. C. So it turns out we can't use this equation over here because we have multiple unknowns. And so we're gonna have to use a different equation. The only other equation that has efficient or the work in it is going to be the efficiency equation. So we can't use this. But it turns out we're gonna have to use this. So our efficiency equation, remember this is a Carnot engine is equal to the work done divided by the heat that's taken in from the hot reservoir, right? It's how much you get out versus how much you paid into this heat engine. However, because this is a Carnot engine, we also have one more equation. We have that. This is one minus TC over th if you look through the variables here, what's going on by the way? This w here is the work done by the engine. That's not to confuse you by w lift. Those are not the same thing. Okay, so what's going on here is that if you look through your variables, we have Q. H, we have T. C. And th here. So what I can do is I can rearrange this equation, I can say the work that's done by the engine. Each cycle is equal to Q. H. Times the efficiency. And because I don't have what the efficiency is, but I can solve it by using this part of the equation. In other words, it's Q. H times one minus TC over th this equation here gives me the work that's done each cycle. So basically all I have to do is just plug some numbers in. So I've got the work done by the engine is in equal to Q. H. Which is 25 times one minus TC over th T. C. here is to 73 And th is 455. So when you work this out, what you're gonna get here is that the work done? Each cycle is 10 jewels. So if you come up here and this kind of makes sense here because you have 25 jewels and you have a work that's done, it's 10, it should be a little bit less than the heat done uh than the heat transferred from the hot reservoir. So what happens here is now that we've figured out the work that's done by the engine. Each cycle, we can now just plug this into our last equation over here, that's the 10 jewels. So all we have to do is we just do n equals the work that's required to lift this mass here divided by the work that's done by the engine. In other words, it's 490 jewels, but each each, each cycle the engine does 10 jewels of work. So in other words, what you're going to need here is you're gonna need 49 cycles of this heat engine running in order to lift this mass by five by five m. Alright, so hopefully all that stuff made sense. Let me know if you guys have any questions in the comments. All right, that's it for this one.

5

Problem

Problem

Your friend claims they have a design for a reversible heat engine that can operate between the freezing and boiling temperatures of water that has an efficiency of 30%. Is this possible?

A

No

B

Yes

C

Not enough information

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